ON THE EXISTENCE OF POSITIVE SOLUTIONS OF HIGHER ORDER DIFFERENCE EQUATIONS. P. J. Y. Wong R. P. Agarwal. 1. Introduction

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1 Topological Methods in Nonlinear Analysis Journal of the Juliusz Schauder Center Volume 10, 1997, ON THE EXISTENCE OF POSITIVE SOLUTIONS OF HIGHER ORDER DIFFERENCE EQUATIONS P. J. Y. Wong R. P. Agarwal 1. Introduction Let a, b (b > a) be integers. We shall denote [a, b] = a, a + 1,..., b}. The notation of all other intervals will carry its stard meaning, e.g. [0, ) denotes the set of nonnegative real numbers. Also, the symbol i denotes the ith forward difference operator with stepsize 1. In this paper we shall consider the n-th order difference equation (1.1) n y + Q(k, y, y,..., n 2 y) = P (k, y, y,..., n 1 y), k [0, N] satisfying the boundary conditions (1.2) (1.3) (1.4) i y(0) = 0, 0 i n 3, α n 2 y(0) β n 1 y(0) = 0, γ n 2 y(n + 1) + δ n 1 y(n + 1) = 0, where n 2, N( n 1) is a fixed positive integer, α, β, γ δ are constants so that (1.5) ρ = αγ(n + 1) + αδ + βγ > Mathematics Subject Classification. 34B15. Key words phrases. Positive solutions, difference equations. c 1997 Juliusz Schauder Center for Nonlinear Studies 339

2 340 P. J. Y. Wong R. P. Agarwal (1.6) α > 0, γ > 0, β 0, δ γ. Further, we assume that there exist functions f : [0, ) [0, ) p, p 1, q, q 1 : [0, N] R such that (i) uf(u) 0 for all u 0, (ii) for u 0, q(k) Q(k, u, u 1,..., u n 2 ) f(u) p(k) P (k, u, u 1,..., u n 1 ) f(u) q 1 (k), p 1 (k), (iii) p 1 (k) is not identical to q(k) p 1 (k) q(k), k [0, N]. We shall give an existence result for positive solutions of the boundary value problem (1.1) (1.4), assuming that f is either superlinear or sublinear. No monotonicity assumption on f is required. To be precise, we introduce the notation f(u) f 0 = lim u 0 u, f f(u) = lim u u. Function f is said to be superlinear if f 0 = 0, f =, f is sublinear provided f 0 =, f = 0. By a positive solution y of (1.1) (1.4), we mean y : [0, N + n] R, y satisfies (1.1) on [0, N], y fulfills (1.2) (1.4), y is nonnegative on [0, N + n], positive on [n 1, N + n 2]. The motivation for the present work stems from many recent investigations. In fact, applications of (1.1) (1.4) their continuous version have been made to singular boundary value problems by Agarwal Wong [2], [15]. Other particular cases of (1.1) (1.4) their continuous analogs have also been the subject matter of several recent publications on singular boundary value problems (e.g. see [1], [5], [10] [12] the references cited therein). In the special case where n = 2, the continuous version of (1.1) (1.4) arises in applications involving nonlinear elliptic problems in annular regions, for this we refer to [3], [4], [9], [14]. In all these applications, it is frequent that only positive solutions are useful. We are particularly motivated by the work of [6] [8], our result is a generalization extension of theirs to a discrete case. The plan of this paper is as follows. In Section 2 we shall state a fixed point theorem due to Krasnosel skiĭ [13], present some properties of certain Green s function which will be used later. In Section 3, we provide an appropriate Banach space a cone so that the fixed point theorem from [13] may be applied to yield a positive solution for (1.1) (1.4).

3 On the Existence of Positive Solutions Preliminaries Theorem 2.1. ([13]) Let B be a Banach space, let C B be a cone in B. Assume that Ω 1, Ω 2 are open subsets of B with 0 Ω 1, Ω 1 Ω 2, let S : C (Ω 2 \Ω 1 ) C be a completely continuous operator such that, either (a) Sy y, y C Ω 1 Sy y, y C Ω 2 or (b) Sy y, y C Ω 1 Sy y, y C Ω 2. Then S has a fixed point in C (Ω 2 \Ω 1 ). To apply Theorem 2.1 in Section 3, we need a mapping whose kernel g(i, j) is the Green s function of the boundary value problem It can be verified that n y = 0, i y(0) = 0, 0 i n 3, α n 2 y(0) β n 1 y(0) = 0, γ n 2 y(n + 1) + δ n 1 y(n + 1) = 0. G(i, j) = n 2 g(i, j) (w.r.t. i) is the Green s function of the boundary value problem Further, we have 2 w = 0, αw(0) β w(0) = 0, γw(n + 1) + δ w(n + 1) = 0. [β + α(j + 1)][δ + γ(n + 1 i)] j [0, i 1], (2.1) G(i, j) = 1 ρ (β + αi)[δ + γ(n j)] j [i, N]. We observe that conditions (1.5) (1.6) imply that G(i, j) is nonnegative on [0, N + 2] [0, N], positive on [1, N + 1] [0, N]. Lemma 2.1. For (i, j) [1, N] [0, N], we find that (2.2) G(i, j) K G(j, j), where 0 < K < 1 is given by (2.3) K = (β + α)(δ + γ) (β + αn)(δ + γn).

4 342 P. J. Y. Wong R. P. Agarwal Proof. For j [0, i 1], using (2.1), we reduce inequality (2.2) to (2.4) [β + α(j + 1)][δ + γ(n + 1 i)] K(β + αj)[δ + γ(n j)]. For (2.4) to hold true, it is sufficient that K satisfies min [β + α(j + 1)][δ + γ(n + 1 i)] K max (β + αj)[δ + γ(n j)], (i,j) [1,N] [0,N] j [0,N] which gives or (β + α)[δ + γ(n + 1 N)] K(β + αn)(δ + γn), (2.5) K or (β + α)(δ + γ) (β + αn)(δ + γn). For j [i, N], inequality (2.2) becomes (β + αi)[δ + γ(n j)] K(β + αj)[δ + γ(n j)], Again, it suffices to find K such that β + αi K(β + αj). which provides min (β + αi) K max i [1,N] (2.6) K β + α β + αn. (β + αj), j [0,N] Taking the intersection of (2.5) (2.6), we immediately get (2.3). Lemma 2.2. For (i, j) [0, N + 2] [0, N], we find that (2.7) G(i, j) L G(j, j), where L > 1 is given by (2.8) L = (β + α)/β β > 0, 2 β = 0. Proof. In the case where j [i, N], from (2.1) it is clear that we may take L = 1 in (2.7). For j [0, i 1], (2.7) is the same as (2.9) [β + α(j + 1)][δ + γ(n + 1 i)] L(β + αj)[δ + γ(n j)]. For (2.9) to hold true, it is sufficient that L satisfies (2.10) [β + α(j + 1)][δ + γ(n j)] L(β + αj)[δ + γ(n j)]

5 On the Existence of Positive Solutions 343 where we have used the fact that 1 i j. If β 0, (2.10) leads to (2.11) L max j [0,N] If β = 0, (2.10) provides β + α(j + 1) β + αj = β + α β. (2.12) L max j [1,N] (j + 1)/j = 2. Expression (2.8) follows immediately from (2.11) (2.12). We shall need the following notations in Section 3. For a nonnegative y( B) which is not identically zero on [0, N], we denote θ = Γ = G(l, l)[q 1 (l) p(l)]f(y(l)) G(l, l)[q(l) p 1 (l)]f(y(l)). In view of (i) (iii), it is clear that θ Γ > 0. Further, we define the constant It is noted that 0 < ξ < 1. ξ = KΓ/Lθ. Let 3. Main results B = y : [0, N + n] R i y(0) = 0, 0 i n 3}, be the Banach space with norm y = max k [0,N+2] n 2 y(k) let C = y B n 2 y(k) be nonnegative is not identically zero We note that C is a cone in B. on [0, N + 2]; Lemma 3.1. Let y B. For 0 i n 3, we find that (3.1) i y(k) k(n 2 i) (n 2 i)! In particular, min k [1,N] n 2 y(k) ξ y }. y, k [0, N + n i]. (3.2) y(k) (N + n)(n 2) (n 2)! y, k [0, N + n].

6 344 P. J. Y. Wong R. P. Agarwal Proof. For y B, we see that which implies n 3 y(k) = n 2 y(l), k [0, N + 3], (3.3) n 3 y(k) k y, k [0, N + 3]. Next, since on using (3.3) we get n 4 y(k) = n 3 y(l), k [0, N + 4], n 4 y(k) l y = k(2) 2! Continuing in the same manner we obtain (3.2). y, k [0, N + 4]. Lemma 3.2. Let y C. For 0 i n 3, we find that (3.4) i y(k) 0, k [0, N + n i], (3.5) i y(k) In particular, (k 1)(n 2 i) (n 2 i)! ξ y, k [1, N + n 2 i]. (3.6) y(k) ξ y, k [n 1, N + n 2]. Proof. Inequality (3.4) is obvious because of the fact that i y(k) = i+1 y(l), k [0, N + n i], 0 i n 3. To prove (3.5), we note that (3.7) n 3 y(k) = n 2 y(l) ξ y = (k 1)ξ y, k [1, N + 1]. Next, using (3.7) we find that l=1 n 4 y(k) = n 3 y(l) (l 1)ξ y = l=1 (k 1)(2) 2! ξ y, for k [1, N +2]. Continuing the process we obtain (3.5). Inequality (3.6) follows immediately from (3.5) when we take i = 0 substitute k = n 1 in the right h side of (3.5).

7 On the Existence of Positive Solutions 345 Remark 3.1. If y C is a solution of (1.1) (1.4), then (3.4) (3.6) imply that y is a positive solution of (1.1) (1.4). To obtain a positive solution of (1.1) (1.4), we shall seek a fixed point of an operator S : C B (3.8) Sy(k) = g(k, l)[q(l, y, y,..., n 2 y) P (l, y, y,..., n 1 y)], for k [0, N + n] in the cone C. It follows that n 2 Sy(k) = G(k, l)[q(l, y, y,..., n 2 y) P (l, y, y,..., n 1 y)], for k [0, N + 2], in view of condition (ii), we get for k [0, N + 2], (3.9) G(k, l)[q(l) p 1 (l)]f(y(l)) n 2 Sy(k) G(k, l)[q 1 (l) p(l)]f(y(l)). Theorem 3.1. Suppose that (i) (iii) hold. If (a) f is superlinear, i.e., f 0 = 0, f = or (b) f is sublinear, i.e., f 0 =, f = 0, then (1.1) (1.4) has a solution in C. Proof. First we shall show that the operator S : C B defined in (3.8) maps C into itself. For this, let y C. Then, from (3.9) (iii) we find (3.10) n 2 Sy(k) G(k, l)[q(l) p 1 (l)]f(y(l)) 0, k [0, N + 2]. Further, it follows from (3.9) Lemma 2.2 that n 2 Sy(k) for k [0, N + 2]. Therefore, (3.11) Sy L L G(k, l)[q 1 (l) p(l)]f(y(l)) G(l, l)[q 1 (l) p(l)]f(y(l)), G(l, l)[q 1 (l) p(l)]f(y(l)) = Lθ.

8 346 P. J. Y. Wong R. P. Agarwal Now, using (3.9), Lemma 2.1 (3.11) we find for k [1, N], Hence, n 2 Sy(k) G(k, l)[q(l) p 1 (l)]f(y(l)) K G(l, l)[q(l) p 1 (l)]f(y(l)) = KΓ ξ Sy. (3.12) min k [1,N] n 2 Sy(k) ξ Sy. It follows from (3.10) (3.12) that S(C) C. Also, stard arguments yield that S is completely continuous. (a) Suppose that f is superlinear. Since f 0 = 0, we may choose a 1 > 0 such that f(u) εu for 0 < u a 1, where ε > 0 satisfies (3.13) Lε(N + n) (n 2) (n 2)! G(l, l)[q 1 (l) p(l)] 1. Let y C be such that y = a 1 (n 2)!/(N + n) (n 2). Then, from (3.2), we have y(k) a 1, k [0, N + n]. Hence, applying (3.9), Lemma 2.2, (3.2) (3.13) successively gives for k [0, N + 2], Consequently, n 2 Sy(k) L G(k, l)[q 1 (l) p(l)]f(y(l)) Lε Lε G(l, l)[q 1 (l) p(l)]f(y(l)) G(l, l)[q 1 (l) p(l)]y(l) (N + n)(n 2) G(l, l)[q 1 (l) p(l)] y y. (n 2)! (3.14) Sy y. If we set then (3.14) holds for y C Ω 1. } Ω 1 = y B y < a 1(n 2)!, (N + n) (n 2)

9 On the Existence of Positive Solutions 347 Next, since f =, we may choose a 2 > 0 such that f(u) Mu for u a 2, where M > 0 satisfies (3.15) ξm G(n 1, l)[q(l) p 1 (l)] 1. Let a 2 = max 2 a 1(n 1)! (N + n), 1 } (n 2) ξ a 2, let y C be such that y = a 2. Then, from (3.6) we have y(k) ε y ξ 1 ξ a 2 = a 2, k [n 1, N + n 2]. Hence, f(y(k)) My(k) for k [n 1, N + n 2]. In view of (3.9), (3.6) (3.15), we find Therefore, n 2 Sy(n 1) G(n 1, l)[q(l) p 1 (l)]f(y(l)) M M G(n 1, l)[q(l) p 1 (l)]f(y(l)) (3.16) Sy y. If we set G(n 1, l)[q(l) p 1 (l)]y(l) G(n 1, l)[q(l) p 1 (l)]ξ y y. Ω 2 = y B y < a2 }, then (3.16) holds for y C Ω 2. In view of (3.14) (3.16), it follows from Theorem 2.1 that S has fixed point y C (Ω 2 \ Ω 1 ) such that a 1 (n 2)! (N + n) (n 2) y a 2. This y is a positive solution of (1.1) (1.4). (b) Suppose that f is sublinear. Since f 0 =, there exists a 3 > 0 such that f(u) Mu for 0 < u a 3, where M > 0 satisfies (3.17) ξm G(n 1, l)[q(l) p 1 (l)] 1.

10 348 P. J. Y. Wong R. P. Agarwal Let y C be such that y = a 3 (n 2)!/(N + n) (n 2). Then, from (3.2), we have y(k) a 3, k [0, N +n]. Hence, using (3.9), (3.6) (3.17) successively, we get n 2 Sy(n 1) G(n 1, l)[q(l) p 1 (l)]f(y(l)) M M G(n 1, l)[q(l) p 1 (l)]f(y(l)) G(n 1, l)[q(l) p 1 (l)]y(l) G(n 1, l)[q(l) p 1 (l)]ξ y y, from which inequality (3.16) follows immediately. If we set } Ω 1 = y B y < a 3(n 2)!, (N + n) (n 2) then (3.16) holds for y C Ω 1. Next, in view of f = 0, we may choose a 4 > 0 such that f(u) ε u for u a 4, where ε > 0 satisfies (3.18) Lε(N + n) (n 2) (n 2)! G(l, l)[q 1 (l) p(l)] 1. There are two cases to consider, namely, f is bounded f is unbounded. Case 1. Suppose that f is bounded, i.e., f(u) R, u [0, ) for some R > 0. Let a 4 = max 2a 3, LR(N + n)(n 2) (n 2)! } G(l, l)[q 1 (l) p(l)], let y C be such that y = a 4 (n 2)!/(N + n) (n 2). For k [0, N + 2], from (3.9) Lemma 2.2 we find n 2 Sy(k) G(k, l)[q 1 (l) p(l)]f(y(l)) R G(k, l)[q 1 (l) p(l)] LR G(l, l)[q 1 (l) p(l)] a 4(n 2)! = y. (N + n) (n 2) Hence, (3.14) holds. Case 2. Suppose that f is unbounded, i.e., there exists } a 3 (n 2)! a 4 > max 2 (N + n), a (n 2) 4

11 On the Existence of Positive Solutions 349 such that f(u) f(a 4 ) for 0 < u a 4. Let y C be such that y = a 4 (n 2)!/(N + n) (n 2). Then, from (3.2) we have y(k) a 4, k [0, N + n]. Hence, applying (3.9), we successively get from Lemma 2.2 (3.18) for k [0, N + 2], n 2 Sy(k) G(k, l)[q 1 (l) p(l)]f(y(l)) L G(l, l)[q 1 (l) p(l)]f(y(l)) L G(l, l)[q 1 (l) p(l)]f(a 4 ) L G(l, l)[q 1 (l) p(l)]εa 4 a 4(n 2)! = y, (N + n) (n 2) from which (3.14) follows immediately. In both Cases 1 2, if we set } Ω 2 = y B y < a 4(n 2)!, (N + n) (n 2) then (3.14) holds for y C Ω 2. Now that we have obtained (3.14) (3.16), it follows from Theorem 2.1 that S has a fixed point y C (Ω 2 \Ω 1 ) such that a 3 (n 2)! (N + n) (n 2) y a 4(n 2)! (N + n) (n 2). This y is a positive solution of (1.1) (1.4). The proof of the theorem is complete. The following two examples illustrate Theorem 3.1. Example 3.1. We consider the boundary value problem 2 y + 2 [k(13 k) + 1] r yr = 0, k [0, 11], 12y(0) y(0) = 0, 12y(12) + 13 y(12) = 0, where r 1. Taking f(y) = y r (which is superlinear if r > 1, sublinear if r < 1), we find Q(k, y) f(y) Hence, we may choose q(k) = = 2 [k(13 k) + 1] r P (k, y, y) f(y) = 0. 1 [k(13 k) + 1] r, q 2 1(k) = [k(13 k) + 1] r, p(k) = p 1 (k) = 0. All conditions of Theorem 3.1 are fulfilled therefore the boundary value problem has a positive solution. One such solution is given by y(k) = k(13 k)+1.

12 350 P. J. Y. Wong R. P. Agarwal Example 3.2. We consider the boundary value problem 3 y + 24k [k(5000 (k 1)(k 6)(k + 1)) + 1] r (y + 1)r = 0, k [0, 10], y(0) = 0, 3 y(0) y(0) = 0, 162 y(11) y(11) = 0, where r < 1. Taking f(y) = (y + 1) r (which is sublinear if r < 1), we find Q(k, y, y) f(y) = 24k [k(5000 (k 1)(k 6)(k + 1)) + 1] r, Hence, we may take P (k, y, y, 2 y) f(y) = 0. q(k) = q 1 (k) = k [k(5000 (k 1)(k 6)(k + 1)) + 1] r, 24k [k(5000 (k 1)(k 6)(k + 1)) + 1] r p(k) = p 1 (k) = 0. Again, all conditions of Theorem 3.1 are satisfied so the boundary value problem has a positive solution. Indeed, y(k) = k[5000 (k 1)(k 6)(k + 1)] is one such solution. References [1] R. P. Agarwal, Difference Equations Inequalities, Marcel Dekker (1992), New York. [2] R. P. Agarwal P. J. Y. Wong, Existence of solutions for singular boundary value problems for higher order differential equations, Rend. Sem. Mat. Fis. Milano 55 (1995), [3] C. Ble, C. V. Coffman M. Marcus, Nonlinear elliptic problems in annular domains, J. Differential Equations 69 (1987), [4] C. Ble M. K. Kwong, Semilinear elliptic problems inannular domains, J. Appl. Math. Phys. 40 (1989), [5] P. W. Eloe J. Henderson, Singular nonlinear boundary value problems for higher order ordinary differential equations, Nonlinear Analysis 17 (1991), [6] P. W. Eloe J. Henderson, Positive solutions for higher order ordinary differential equations, preprint. [7] P. W. Eloe, J. Henderson P. J. Y. Wong, Positive solutions for two-point boundary value problems, In Proceedings of the Dynamic Systems Applications 2, ed. G. S. Ladde M. Sambham, Atlanta, Dynamic Publishers, 1996,

13 On the Existence of Positive Solutions 351 [8] L. H. Erbe H. Wang, On the existence of positive solutions of ordinary differential equations, Proc. Amer. Math. Soc. 120 (1994), [9] X. Garaizar, Existence of positive radial solutions for semilinear elliptic problems in the annulus, J. Differential Eqations 70 (1987), [10] J. A. Gatica, V. Oliker P. Waltman, Singular nonlinear boundary value problems for second-order ordinary differential equations, J. Differential Eqations 79 (1989), [11] J. Henderson, Singular boundary value problems for difference equations, Dynam. Systems Appl. 1 (1992), [12] J. Henderson, Singular boundary value problems for higher order difference equations, In Proceedings of the First World Congress of Nonlinear Analysts, ed. V. Lakshmikantham, Berlin, Walter de Gruyter Co. (1996), [13] M. A. Krasnosel skĭ, Positive Solutions of Operator Equations (1964), Noordhoff, Groningen. [14] H. Wang, On the existence of positive solutions for semilinear elliptic equations in the annulus, J. Differential Eqations 109 (1994), 1 7. [15] P. J. Y. Wong R. P. Agarwal, On the existence of solutions of singular boundary value problems for higher order difference equations, Nonlinear Analysis 28 (1997), Manuscript received March 12, 1996 Patricia J. Y. Wong Division of Mathematics Nanyang Technological University 469, Bukit Tinah Road Singapore , SINGAPORE address: wongjyp@nievax.nie.ac.sg Ravi P. Agarwal Department of Mathematics National University of Singapore 10, Kent Ridge Crescent Singapore , SINGAPORE address: matravip@leonis.nus.edu.sg TMNA : Volume N o 2